Question

examine the following graph of a function modeling damped harmonic motion. find the equation for the function pictured in terms of y and t. assume that a factor of (e^{-t}) provides the desired damping effect and that the graph has no vertical or horizontal shifts.

Explanation:

Step1: Recall the general form of damped - harmonic motion

The general form of a damped - harmonic motion function with no vertical or horizontal shifts and exponential damping factor $e^{-t}$ is $y = Ae^{-t}\sin(bt)$ or $y = Ae^{-t}\cos(bt)$. When $t = 0$, for $y = Ae^{-t}\cos(bt)$, $y(0)=A\cos(0)=A$; for $y = Ae^{-t}\sin(bt)$, $y(0)=A\sin(0) = 0$. Since the graph passes through the point $(0, - 2)$, we use the cosine - based form $y = Ae^{-t}\cos(bt)$. Substituting $t = 0$ and $y=-2$ into $y = Ae^{-t}\cos(bt)$, we get $-2=Ae^{0}\cos(0)$, so $A=-2$.

Step2: Determine the value of $b$

The period of the undamped cosine function (ignoring the damping factor) can be determined from the graph. The distance between two consecutive peaks or troughs gives the period. From the graph, the period $T$ of the undamped cosine - like function is $T = 2$. The formula for the period of $y=\cos(bt)$ is $T=\frac{2\pi}{b}$. Since $T = 2$, we have $2=\frac{2\pi}{b}$, and solving for $b$ gives $b=\pi$.

Answer:

$y=-2e^{-t}\cos(\pi t)$